Answer:
1. Density of Metal = 4.40g/cm³
2. Density of Liquid = 0.77568g/cm³
Explanation:
1.
Given
Initial Volume of Water = Vi = 21.27mL
Volume of Water = VF = 24.78 mL
Mass of Sphere= 15.45 g
Density = Mass/Volume
Change in Volume = 24.78 - 21.27 = 3.51 mL
Density = 15.45/3.51
Density = 4.40g/ml
1ml = 1cm³
So, density = 4.40/cm³
2.
Initial Mass of Beaker = Mb = 32.4257g
Final Mass of Beaker = Mf = 40.1825 g
V = 10 mL
Density = D = ?
Change in Mass =: MF – MI= 40.1825 – 32.4257 = 7.7568 g
Density = Mass/Volume
Density = 7.7568g/10mL
Density = 0.77568g/mL
1mL = 1cm³
So, density = 0.77568g/cm³
Concentration = 0.5 N
volume = 250 ml = 250 cm^3 = 0.250 dm^3
moles = conc. × vol.
= 0.5 × 0.250
= 0.125
mass = moles × relative molecular mass
so relative molecular mass if Na2CO3
= (23 × 2 ) +( 12 × 1 )+ (16 × 3 )
= 46 + 12 + 48
= 106
mass = 0.125 × 106
= 13.25 g
Hope it helped !!
Answer:
1.75M
Explanation:
Given parameters:
Initial volume of acid, HCl = 450mL = 0.45L
Initial concentration = 3.5M
Final volume = 0.9L
Unknown:
Final concentration = ?
Solution:
This is a dilution problem in which a particular concentration is made from the stock of known concentration.
One important approach to solve this problem is to remember that the number of moles in the initial and final solution will always remain the same.
Since we know this;
Number of moles = molarity x volume;
let us find the number of moles of the initial solution;
Number of moles = 3.5 x 0.45 = 1.58moles
Now, to find the new molarity;
Molarity =
Input the parameters;
Molarity = = 1.75M
The transfer of energy that occurs when a force is applied over a distance is WORK.
Hope this helps!